Constants Part A An electron moves at 2.10x106 m/s through a region in which the

Question

Constants Part A An electron moves at 2.10x106 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.60x10-2 T What is the largest possible magnitude of the acceleration of the electron due to the magnetic field? Submit Request Answer Part B What is the smallest possible magnitude of the acceleration of the electron due to the magnetic field? Submit equest Answer Part C If the actual acceleration of the electron isof the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field? Submit Request Answer

Explanation / Answer

Fb = q (v x B) = q v B sin(theta)

a = Fb/m = (q v B/m) sin(theta)

a = (1.6 x 10^-19 x 2.10 x 10^6 x 7.60 x 10^-2 / (9.109 x 10^-31)) sin(theta)

a = (2.81 x 10^16) sin(theta)

(A) a will be max when theta = 90

a_max = 2.81 x 10^16 m/s^2

(B) a-min when theta = 0 then a = 0

(C) then sin(theta) = 1/4

theta = 14.5 deg

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---